Is ${441853}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {441853}= &&{4}\cdot100000+ \\&&{4}\cdot10000+ \\&&{1}\cdot1000+ \\&&{8}\cdot100+ \\&&{5}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {441853}= &&{4}(99999+1)+ \\&&{4}(9999+1)+ \\&&{1}(999+1)+ \\&&{8}(99+1)+ \\&&{5}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {441853}= &&\gray{4\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{5\cdot9}+ \\&& {4}+{4}+{1}+{8}+{5}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${441853}$ is divisible by $9$ if ${ 4}+{4}+{1}+{8}+{5}+{3}$ is divisible by $9$ Add the digits of ${441853}$ $ {4}+{4}+{1}+{8}+{5}+{3} = {25} $ If ${25}$ is divisible by $9$ , then ${441853}$ must also be divisible by $9$ ${25}$ is not divisible by $9$, therefore ${441853}$ must not be divisible by $9$.